System of ODE - Solution

Question

I have a system of ODE to solve

$$ A_{5 \times 5}\ddot{q}(t)_{5 \times 1}+ B_{5 \times 5}\dot{q}(t)_{5 \times 1}+ C_{5 \times 1} =0\tag 1$$

Given Data

1. $A,B,C$ are constants.We know what is $q(0),q'(0)$
2. $\dot{q}$ means derivative of vector $q$ w.r.t $t$

Question(2 Nos)

1. How can we solve this ODE in closed form?.
2. How can we solve this ODE by numerical methods?.

NB:: Method using reducing order can also be considered

Answer 1

Let $v = \dot{q}$, so your system is $$ A \dot{v} + B v + C = 0 $$ I suppose you're given the initial conditions $v(0)$ (not just $q(0)$). Of course once $v(t)$ is found, integration (with the initial conditions $q(0)$) gives you $q(t)$.

If $A$ is invertible, your system is equivalent to $$\dot{v} + A^{-1} B v + A^{-1} C = 0$$ If $A$ is not invertible, use Gauss-Jordan elimination. Possibly re-ordering the variables, you get an invertible matrix $E$ such that $EA$ is of the block form

$$ \pmatrix{I & A_{2}\cr 0 & 0\cr}$$

Multiplying your system by $E$, and writing $v = \pmatrix{x\cr y\cr}$, the system becomes

$$ \pmatrix{I & A_2 \cr 0 & 0\cr} \pmatrix{\dot{x}\cr \dot{y}\cr} + \pmatrix{B_1 & B_2\cr B_3 & B_4\cr} \pmatrix{x\cr y\cr} + \pmatrix{C_1\cr C_2\cr} = 0 $$

i.e. $$ \eqalign{\dot{x} + A_2 \dot{y} + B_1 x + B_2 y + C_1 &= 0\cr B_3 x + B_4 y + C_2 &= 0\cr} \tag{1}$$

For simplicity, let's suppose $B_4$ is invertible, so we can solve the second equation of (1) as

$$ y = -B_4^{-1} B_3 x - B_4^{-1} C_2$$

This must be true for the initial conditions $x(0), y(0)$, and then we must have $$ \dot{y} = - B_4^{-1} B_3 \dot{x}$$

Substitute these into the first equation of (1), and you get an equation for $\dot{x}$ and $x$:

$$ (I - A_2 B_4^{-1} B_3) \dot{x} + (B_1 - B_2 B_4^{-1} B_3) x + C_1 - B_2 B_4^{-1} C_2 = 0 $$

Of course it's possible that $I - A_2 B_4^{-1} B_3$ is not invertible, in which case you go through this process again...